Groups of Small Orders

We have developed lots of theories about groups. We can see them in practice in the study of groups of small orders.

Useful Lemmas

Lemma. Let $G$ be a finite group in which every element except the identity has order $2$. Then $G$ is abelian, and has order $2^m$ for some integer $m$.

Proof.

For any $g, h \in G$, $gh \in G$ is also of order $2$, therefore $gh = (gh)^{-1} = h^{-1}g^{-1} = hg$ and $G$ is abelian.

Let $H = \set{g_1, g_2, …, g_m}$ be the minimal generating set for $G$, which means $H$ generates $G$ and no proper subset of $H$ can do so. Hence, every $g \in G$ is a power of elements in $H$, as $G$ is abelian, we have

\[g = g_1^{p_1}g_2^{p_2}...{g_m}^{p_m}\]

Suppose $g = g_1^{p_1}g_2^{p_2}…{g_m}^{p_m} = g_1^{q_1}g_2^{q_2}…{g_m}^{q_m}$ where $p_i$ and $q_i$ can only be $0$ or $1$. If $p_i \not = q_i$, we have $g_i = g_1^{r_1}g_2^{r_2}…g_{i-1}^{r_{i-1}}g_{i+1}^{r_{i+1}}…g_m^{r_m}$ which contradicts with the fact that $H$ is the minimal generating set. So the expression is unique and every combination of $p_1, p_2, …, p_m$ is a distinct element in $G$.

Lemma. Let $G$ be a group of even order. Then $G$ contains an element of order $2$.

Proof.

Suppose $G$ has $2n$ elements. By pairing every element in $G$ with its inverse, we will have at least 2 sets having only one element, one of which is $\set{e}$. The element in another set will have order $2$.

Groups of order 3

Theorem. Every group of order $3$ is cyclic.

Proof.

Every group of prime order is cyclic. But we will analyze it from first principles.

Let $G = \set{e, x, y}$ be a group of order $3$. We have $yx \not = x$ and $yx \not = y$ so $yx = e$. By Langrange’s Corollary, $x^3 = e$, so $y = yx^3 = x^2$. Hence, $G = \set{e, x, x^2}$ is a cyclic group.

Groups of order 4

Theorem. Every group of order $6$ is either cyclic or dihedral.

Proof.

By Lagrange, the order of every element is either $1, 2$ or $4$. If there is an element of order $4$, then $G \cong C_4$.

If there is no element of order $4$, all elements other than the identity have order $2$. Therefore, $G$ is abelian. Picking any two elements we have $x^2 = y^2 = e$ and $xy = yx = y^{-1}x$. So, $G \cong D_4 \cong V_4$.

Groups of order 6

Theorem. Every group of order $6$ is either cyclic or dihedral.

Proof.

By Lagrange, the order of every element is either $1, 2, 3$ or $6$. If there is an element of order $6$, then $G \cong C_6$.

If there is no element of order $6$, elements other than the identity can only have orders $2$ or $3$. There must be an element $y$ of order $3$, otherwise every element is of order $2$ and the order of group has to be powers of $2$. $\langle y \rangle$ is a normal subgroup as it has index $2$. Also, there must be an element $x$ of order $2$ as the order of $G$ is even. Therefore, we have $xyx^{-1} \in \langle y \rangle$ and

  • $xyx^{-1} = e$, $y = e$ which is not possible, or

  • $xyx^{-1} = y$, $xy = yx$ so $(xy)^m$ = $x^my^m$ and $xy$ is of order $6$ (can’t be $2$ or $3$) and $G \cong C_6$, or

  • $xyx^{-1} = y^2 = y^{-1}$, so $xy = y^{-1}x$, together with $x^2 = y^3 = e$, $G \cong D_6$.

Groups of order 8

Theorem. Every group of order 8 is either

  • abelian and isomorphic to one of $C_8$, $C_4 \times C_2$, $C_2 \times C_2 \times C_2$, or

  • not abelian and isomorphic to one of $D_8$ or $Q_8$.

Proof.

By Lagrange, the order of every elements is either $1, 2, 4$ or $8$. If there is an element of order $8$, then $G \cong C_8$.

Suppose that the largest order of an element of $G$ is $4$. Choose an element $x$ whose order is $4$ and an element $y$ from $G \setminus \langle x \rangle$. As $\langle x \rangle$ is of index $2$, the cosets $\langle x \rangle \cup \langle x \rangle y = G$. Thus, the elements of $G$ are

\[G = \Set{e, x, x^2, x^3, y, xy, x^2y, x^3y}\]

Consider the element $yx$. $yx \not \in \langle x \rangle$ otherwise $y \in \langle x \rangle$. $yx \not = y$ otherwise $x = e$. $yx \not = x^2y$ otherwise $x = y^{-1}x^2y \implies x^2 = y^{-1}x^4y = e$. Therefore, $yx = xy$ or $yx = x^3y = x^{-1}y$.

Consider the element $y^2$. $y^2 = yy \not \in \langle x \rangle y$ otherwise $y \in \langle x \rangle$. $y^2 \not= x, x^3$ othwerwise $y$ has order $8$. Therefore, $y^2 = e$ or $y^2 = x^2$.

Hence, we have in total four possibilities:

  • $yx = xy$ and $y^2 = e$, the group is abelian, generated by $\Set{x, y}$ and isomorphic to $C_4 \times C_2$.

  • $yx = xy$ and $y^2 = x^2$, the group is abelian, generated by $\Set{x, yx^{-1}}$ as $(yx^{-1})^2 = y^2x^{-2} = e$ and isomorphic to $C_4 \times C_2$.

  • $yx = x^3y$ and $y^2 = e$, the group satisfies $x^4 = y^2 = e$ and $yx = x^{-1}y$ and is isomorphic to $D_8$.

  • $yx = x^3y$ and $y^2 = x^2$, the group satisfies $x^4 = e$, $y^2 = x^2$ and $yx = x^{-1}y$ and is isomorphic to $Q_8$.

Finally, if every element except the identity is of order $2$, then $G$ is abelian. Choose $x, y, z$ from $G \setminus \Set{e}$ such that $xy \not= z$. We have $H = \Set{e, x, y, xy} \cong C_2 \times C_2$ and if $K = \Set{e, z}$, $HK = G$ and $H \cap K = \Set{e}$. By Direct Product Theorem, the group is isomorphic to $C_2 \times C_2 \times C_2$.

Groups of order $p$

Theorem. Suppose $G$ is a finite group of order $p$, where $p$ is prime. Then $G$ is cyclic and the only subgroups are $\set{e}$ and $G$.

Proof.

For any $a \in G$, by Lagrange, $\text{ord}(a) = p$ and generates $G$, so $G$ is cyclic.

For $H \le G$, by Lagrange, $|H|$ divides $|G| = p$, hence $|H| = 1$ or $p$. When $|H| = 1$, $H = \set{e}$. When $|H| = p$, $H = G$. .

Groups of order $2p$

Theorem. Every group of order $2p$, where $p$ is prime, is either cyclic or dihedral.

Proof.

When $p = 2$, the group is of order $4$ and as shown above it is either cyclic or dihedral.

Assume $p > 2$, by Lagrange, the order of every element is either $1, 2, p$ or $2p$. If there is an element of order $2p$, then $G \cong C_{2p}$.

If there is no element of order $2p$, elements other than identity can only have orders $2$ or $p$. There must be an element $y$ of order $p$, otherwise every element is of order $2$ and the order of group has to be powers of $2$. $\langle y \rangle$ is a normal subgroup as it has index $2$. Also, there must be an element $x$ of order $2$ as the order of $G$ is even. Therefore, we have $xyx^{-1} \in \langle y \rangle$ so $xyx^{-1} = y^k$ for $0 \le k \le p - 1$. Furthermore, $y = xy^kx = (xyx)^k = y^{k^2}$ so $k^2 = 1$.

  • $xyx^{-1} = e$, $y = e$ which is not possible, or

  • $xyx^{-1} = y$, $xy = yx$ so $(xy)^m$ = $x^my^m$ and $xy$ is of order $6$ (can’t be $2$ or $3$) and $G \cong C_6$, or

  • $xyx^{-1} = y^2 = y^{-1}$, so $xy = y^{-1}x$, together with $x^2 = y^3 = e$, $G \cong D_6$.

Groups of order $p^n$

Lemma. Every group of order $p^n$, where $p$ is prime, has non-trival centre.

Proof.

Consider the conjuation on $G$ by $G$. $Z(G)$ consists of elements that belong to conjugacy classes of size $1$. As

\[|[g]| = |G| / |C_G(g)|\]

$|[g]]|$ is divisible by $p$. On the other hand, the conjugacy classes parition $G$, so

\[|G| = \sum_{g \in G} |[g]| = \text{# of conjugacy classes of size 1} + \sum_{|[g]| > 1} |[g]|\]

Thus, we have number of conjugacy classes of size $1$ divisible by $p$ and therefore at least $p - 1$ of them other than $[e]$. Hence, $Z(G)$ is non-trival.

Lemma. If $G / Z(G)$ is cyclic, $G = Z(G)$.

Proof.

If $G / Z(G)$ is cyclic, pick $g \in G$ such that $G / Z(G) = \Set{g^nZ(G) : n \in \mathbb{Z}}$. Every element in $G$ is therefore of the form $g^n z$ for some integer $n$ and $z \in Z(G)$ and commutes with each other. Hence, $G = Z(G)$.

Theorem. Every group of order $p^2$, where $p$ is prime, is abelian and isomorphic to $C_{p^2}$ or $C_p \times C_p$.

Proof.

The order of the centre $Z(G)$ can be $1, p, p^2$. However, $1$ is not possible as the $Z(G)$ is non-trival. Also, $p$ is not possible as $G / Z(G)$ will be cyclic and $G = Z(G)$. Hence, $Z(G)$ has order $p^2$ and therefore $G = Z(G)$ and is abelian.

From that, if $G$ has an element of order $p^2$, then $G \cong C_{p^2}$. If $G$ has only element of order $p$, then $G \cong C_p \times C_p$.

References